5Symmetry methods in PDEs
II Integrable Systems
5.4 Jets and prolongations
This is all nice, but we still need to find a way to get Lie point symmetries. So
far, we have just found them by divine inspiration, which is not particularly
helpful. In general. Is there a more systematic way of finding Lie symmetries?
We can start by looking at the trivial case — a 0th order ODE
∆[x, u] = 0.
Then we know g
ε
: (x, u) 7→ (˜x, ˜u) is a Lie point symmetry if
∆[x, u] = 0 =⇒ ∆[˜x, ˜u] = ∆[g
ε
(x, u)] = 0.
Can we reduce this to a statement about the generator of g
ε
? Here we need to
assume that ∆ is of maximal rank, i.e. the matrix of derivatives
∂∆
j
∂y
i
is of maximal rank, where the
y
i
runs over
x, u
, and in general all coordinates.
So for example, the following theory will not work if, say ∆[
x, u
] =
x
2
. Assuming
∆ is indeed of maximal rank, it is an exercise on the example sheet to see that if
V is the generator of g
ε
, then g
ε
is a Lie point symmetry iff
∆ = 0 =⇒ V (∆) = 0.
This essentially says that the flow doesn’t change ∆ iff the derivative of ∆
along
V
is constantly zero, which makes sense. Here we are thinking of
V
as a
differential operator. We call this constraint an on-shell condition, because we
only impose it whenever ∆ = 0 is satisfied, instead of at all points.
This equivalent statement is very easy! This is just an algebraic equation for
the coefficients of V , and it is in general very easy to solve!
However, as you may have noticed, these aren’t really ODE’s. They are just
equations. So how do we generalize this to
N ≥
1 order ODE’s? Consider a
general vector field
V (x, u) = ξ(x, u)
∂
∂x
+ η(x, u)
∂
∂u
.
This only knows what to do to
x
and
u
. But if we know how
x
and
u
change, we
should also know how
u
x
, u
xx
etc. change. Indeed this is true, and extending the
action of V to the derivatives is known as the prolongation of the vector field.
We start with a concrete example.
Example. Consider a 1 parameter group of transformations
g
ε
: (x, u) 7→ (e
ε
x, e
−ε
u) = (˜x, ˜u)
with generator
V = x
∂
∂x
− u
∂
∂u
.
This induces a transformation
(x, u, u
x
) 7→ (˜x, ˜u, ˜u
˜x
)
By the chain rule, we know
d˜u
d˜x
=
d˜u/dx
d˜x/dx
= e
−2ε
u
x
.
So in fact
(˜x, ˜u, ˜u
˜x
) ≡ (e
ε
x, e
−ε
u, e
−2ε
u
x
).
If we call (
x, u
) coordinates for the base space, then we call the extended
system (
x, u, u
x
) coordinates for the first jet space. Given any function
u
=
u
(
x
),
we will get a point (x, u, u
x
) in the jet space for each x.
What we’ve just seen is that a one-parameter group of transformation of the
base space induces a one-parameter group of transformation of the first jet space.
This is known as the prolongation, written
pr
(1)
g
ε
: (x, u, u
x
) 7→ (˜x, ˜u, ˜u
˜x
) = (e
ε
x, e
−ε
u, e
−2ε
u
x
).
One might find it a bit strange to call
u
x
a coordinate. If we don’t like doing
that, we can just replace
u
x
with a different symbol
p
1
. If we have the
n
th
derivative, we replace the nth derivative with p
n
.
Since we have a one-parameter group of transformations, we can write down
the generator. We see that pr
(1)
g
ε
is generated by
pr
(1)
V = x
∂
∂x
− u
∂
∂u
− 2u
x
∂
∂u
x
.
This is called the first prolongation of V .
Of course, we can keep on going. Similarly,
pr
(2)
g
ε
acts on the second jet
space which has coordinates (x, u, u
x
, u
xx
). In this case, we have
pr
(2)
g
ε
(x, u, u
x
, u
xx
) 7→ (˜x, ˜u, ˜u
˜x˜x
) ≡ (e
ε
x, e
−ε
u, e
−2ε
u
x
, e
−3ε
u
xx
).
This is then generated by
pr
(2)
V = x
∂
∂x
− u
∂
∂u
− 2u
x
∂
∂u
x
− 3u
xx
∂
∂u
xx
.
Note that we don’t have to recompute all terms. The
x, u, u
x
terms did not
change, so we only need to check what happens to ˜u
˜x˜x
.
We can now think of an nth order ODE
∆[x, u, u
x
, ··· , u
(n)
] = 0
as an algebraic equation on the
n
th jet space. Of course, this is not just an
arbitrary algebraic equation. We will only consider solutions in the
n
th jet space
that come from some function
u
=
u
(
x
). Similarly, we only consider symmetries
on the
n
th jet space that come from the prolongation of some transformation on
the base space.
With that restriction in mind, we have effectively dressed up our problem
into an algebraic problem, just like the case of ∆[
x, u
] = 0 we discussed at the
beginning. Then g
ε
: (x, u) 7→ (˜x, ˜u) is a Lie point symmetry if
∆[˜x, ˜u, ˜u
˜x
, . . . , ˜u
(n)
] = 0
when ∆ = 0. Or equivalently, we need
∆[pr
(n)
g
ε
(x, u, . . . , u
(n)
)] = 0
when ∆ = 0. This is just a one-parameter group of transformations on a
huge coordinate system on the jet space. Thinking of all
x, u, ··· , u
(n)
as just
independent coordinates, we can rewrite it in terms of vector fields. (Assuming
maximal rank) this is equivalent to asking for
pr
(n)
V (∆) = 0.
This results in an overdetermined system of differential equations for (
ξ, η
),
where
V (x, u) = ξ(x, u)
∂
∂x
+ η(x, u)
∂
∂u
.
Now in order to actually use this, we need to be able to compute the
n
th
prolongation of an arbitrary vector field. This is what we are going to do next.
Note that if we tried to compute the prolongation of the action of the Lie
group, then it would be horrendous. However, what we actually need to compute
is the prolongation of the vector field, which is about the Lie algebra. This makes
it much nicer.
We can write
g
ε
(x, u) = (˜x, ˜u) = (x + εξ(x, u), uε + η(x, u)) + o(ε).
We know the nth prolongation of V must be of the form
pr
(n)
V = V +
n
X
k=1
η
k
∂
∂u
(k)
,
where we have to find out what η
k
is. Then we know η
k
will satisfy
g
ε
(x, u, ··· , u
(n)
) = (˜x, ˜u, ··· , ˜u
(n)
)
= (x + εξ, u + εη, u
x
+ η
1
, ··· , u
(n)
+ εη
n
) + o(ε).
To find η
1
, we use the contact condition
d˜u =
d˜u
d˜x
d˜x = ˜u
˜x
d˜x.
We now use the fact that
˜x = x + εξ(x, u) + o(ε)
˜u = x + εη(x, u) + o(ε).
Substituting in, we have
du + εdη = ˜u
˜x
(dx + εdξ) + o(ε).
We want to write everything in terms of dx. We have
du = u
x
dx
dη =
∂η
∂x
dx +
∂η
∂u
du
=
∂η
∂x
+ u
x
∂η
∂u
dx
= D
x
η dx,
where D
x
is the total derivative
D
x
=
∂
∂x
+ u
x
∂
∂u
+ u
xx
∂
∂u
x
+ ··· .
We similarly have
dξ = D
x
ξdx.
So substituting in, we have
(u
x
+ εD
x
η)dx = ˜u
˜x
(1 + εD
x
ξ)dx + o(ε).
This implies that
˜u
˜x
=
u
x
+ εD
x
η
1 + εD
x
ξ
+ o(ε)
= (u
x
+ εD
x
η)(1 − εD
x
ξ) + o(ε)
= u
x
+ ε(D
x
η −u
x
D
x
ξ) + o(ε).
So we have
η
1
= D
x
η −u
x
D
x
η.
Now building up η
k
recursively, we use the contact condition
d˜u
(k)
=
d˜u
(k)
d˜x
d˜x = ˜u
(k+1)
d˜x.
We use
˜u
(k)
= u
(k)
+ εη
k
+ o(ε)
˜x = x + εξ + o(ε).
Substituting that back in, we get
(u
(k+1)
+ εD
x
η
k
) dx = ˜u
(k+1)
(1 + εD
x
ξ)dx + o(ε).
So we get
˜u
(k+1)
= (u
(k+1)
+ εD
x
η
k
)(1 − εD
x
ξ) + o(ε)
= u
(k+1)
+ ε(D
x
η
k
− u
(k+1)
D
x
ξ) + o(ε).
So we know
η
k+1
= D
x
η
k
− u
(k+1)
D
x
ξ.
So we know
Proposition (Prolongation formula). Let
V (x, u) = ξ(x, u)
∂
∂x
+ η(x, u)
∂
∂u
.
Then we have
pr
(n)
V = V +
n
X
k=1
η
k
∂
∂u
(k)
,
where
η
0
= η(x, u)
η
k+1
= D
x
η
k
− u
(k+1)
D
x
ξ.
Example. For
g
ε
: (x, u) 7→ (e
ε
x, e
−ε
u),
we have
V = x
∂
∂x
+ (−u)
∂
∂u
.
So we have
ξ(x, u) = x
η(x, u) = −u.
So by the prolongation formula, we have
pr
(1)
V = V + η
1
∂
∂u
x
,
where
η
x
= D
x
(−u) − u
x
D
x
(x) = −2u
x
,
in agreement with what we had earlier!
In the last example sheet, we will derive an analogous prolongation formula
for PDEs.